Question

An air filled parallel plate capacitor has a capacity \(2\ \text{pF}\). The separation of the plates is doubled and the interspace between the plates is filled with dielectric material, then the capacity is increased to \(6\ \text{pF}\). The dielectric constant of the material is

• A. 3
• B. 6
• C. 2
• D. 4

Hint:

When separation changes, first find the capacitance without the dielectric: doubling \(d\) alone would halve the capacitance. Then multiply by \(K\) to get the final value. Set up the ratio \(\frac{C_2}{C_1} = \frac{K}{2}\) to solve quickly.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
An air‑filled parallel plate capacitor has capacitance \(C_1 = 2\ \text{pF}\). The plate separation is then doubled (\(d' = 2d\)) and a dielectric material fills the space. The new capacitance becomes \(C_2 = 6\ \text{pF}\). We need the dielectric constant \(K\) of the material.

Step 2: Key Formula or Approach:
For a parallel plate capacitor: \[ C = \frac{\varepsilon_0 A}{d} \quad (\text{air filled}). \] When a dielectric of constant \(K\) is inserted and the separation changes: \[ C' = \frac{K \varepsilon_0 A}{d'}. \]

Step 3: Detailed Explanation:
Initially, with air (\(K=1\)): \[ C_1 = \frac{\varepsilon_0 A}{d} = 2\ \text{pF}. \] After doubling the separation and adding dielectric: \[ C_2 = \frac{K \varepsilon_0 A}{2d} = \frac{K}{2} \cdot \frac{\varepsilon_0 A}{d} = \frac{K}{2} \cdot C_1. \] Substitute the given values: \[ 6 = \frac{K}{2} \times 2 \quad \Rightarrow \quad 6 = K. \] Thus the dielectric constant \(K = 6\).

Step 4: Final Answer:
The dielectric constant is 6, which is option (B).